x =12 and y =9
3x -18 = 2y or 3x -2y = 18
5x-6y = 6
using elimination method
in order make coefficient of y in both equation same with different sign ,
we multiply equation 1 by -3 and adding the obtained to equation 2
-3(3x-2y) = 18x-3
-9x +6y = -54
5x -6y = 6
-4x = -48
on plugging the value of y in equation (1) ,we get
3(12) -18 = 2y
36 -18 = 2y
2y = 18
y = 18/2 = 9
x = 12 and y = 9
x = [-8, 8] + 5/7·([-15, -13] - [-8, 8]) = [-13, -7]
d. 12, 6, 7
for the triangle to have non-zero area, the sum of the shortest two sides must exceed the longest side.
a: 3 + 7 = 10 . . triangle has zero area
b: 4 + 5 < 10 . . sides don't meet
c: 6 + 8 = 14 . . triangle has zero area
d: 6 + 7 > 12 . . these form a triangle with non-zero area.
for #2, another way to word this question is: for which of the following trig functions is π/2 a solution? well, go through them one by one. if you plug π/2 into sinθ, you get 1. this means that when x is π/2, y is 1. try and visualize that. when y is 1, that means you moved off the x-axis; so y = sinθ is not one of those functions that cross the x-axis at θ = π/2. go through the rest of them. for y = cos(π/2), you get 0. at θ = π/2, this function crosses the x-axis. for y = tanθ, your result is undefined, so that doesn't work. keep going through them. you should see that y = secθ is undefined, y = cscθ returns 1, and y = cotθ returns 0. if you have a calculator that can handle trig functions, just plug π/2 into every one of them and check off the ones that give you zero. graphically, if the y-value is 0, the function is touching/crossing the x-axis.
think about what y = secθ really means. it's actually y = 1/(cosθ), right? so what makes a fraction undefined? a fraction is undefined when the denominator is 0 because in mathematics, you can't divide by zero. calculators give you an error. so the real question here is, when is cosθ = 0? again, you can use a calculator here, but a unit circle would be more . cosθ = π/2, like we just saw in the previous problem, and it's zero again 180 degrees later at 3π/2. now read the answer choices.
all multiples of pi? well, our answer looked like π/2, so you can skip the first two choices and move to the last two. all multiples of π/2? imagine there's a constant next to π, say cπ/2 where c is any number. if we put an even number there, 2 will cut that number in half. imagine c = 4. then cπ/2 = 2π. our two answers were π/2 and 3π/2, so an even multiple won't work for us; we need the odd multiples only. in our answers, π/2 and 3π/2, c = 1 and c = 3. those are both odd numbers, and that's how you know you only need the "odd multiples of π/2" for question 3.